Solved Problems on Reciprocal Cones

Squaring the given equation twice, one obtains $$f^2x^2+g^2y^2+h^2z^2-2fgxy-2ghyz-2hfxz=0. \tag{1}$$ This is a homogeneous equation of second degree. The determinant of its coefficients is $$\Delta= \begin{vmatrix} f^2 & -fg & -hf\\ -fg & g^2 & -gh\\ -hf & -gh & h^2 \end{vmatrix} =-4(fgh)^2\neq 0.$$ Hence, (1) represents a cone with vertex at the origin. The reciprocal cone is $$Ax^2+By^2+Cz^2+2Fyz+2Gzx+2Hxy=0,$$ where $A,B,C,F,G,H$ are the cofactors of $f^2,g^2,h^2,-gh,-hf,-fg$ in $\Delta$. Thus, $$A=B=C=0,\quad F=2f^2gh,\quad G=2g^2hf,\quad H=2h^2fg.$$ Therefore, the reciprocal cone is $$f^2ghyz+g^2hzx+h^2fgxy=0,$$ or equivalently, $$fyz+gzx+hxy=0.$$ This cone contains the coordinate axes, while the original cone touches all the coordinate planes. $\blacksquare$

Let $P(x,y,z)$ be a general point on the cone. It is assumed that the generator through $P$ meets the base plane at $(x_1,y_1,0)$. Then, $$x_1^2+y_1^2-2cx_1=0. \quad (1)$$ The equations of the generator are $$\frac{x}{x_1}=\frac{y}{y_1}=\frac{z-d}{-d}. \quad (2)$$ From (2), one obtains $$x_1=-\frac{dx}{z-d}, \qquad y_1=-\frac{dy}{z-d}.$$ Substituting in (1), $$\frac{d^2x^2}{(z-d)^2}+\frac{d^2y^2}{(z-d)^2}+2c\frac{dx}{z-d}=0,$$ which simplifies to $$dx^2+dy^2+2cx(z-d)=0. \quad (3)$$ Equation (3) represents the given cone with vertex $(0,0,d)$. A translation of axes is now applied by taking $$x=X,\quad y=Y,\quad z=Z+d.$$ Thus, the vertex is shifted to the origin and the equation of the cone becomes $$dX^2+dY^2+2cXZ=0.$$ The reciprocal cone of $$aX^2+bY^2+cZ^2+2fYZ+2gZX+2hXY=0$$ is known to be $$AX^2+BY^2+CZ^2+2FYZ+2GZX+2HXY=0,$$ where $A,B,C,F,G,H$ are the cofactors of $a,b,c,f,g,h$ respectively in the determinant $$\begin{vmatrix} a & h & g\\ h & b & f\\ g & f & c \end{vmatrix}.$$ In the present case, the determinant is $$\begin{vmatrix} d & 0 & c\\ 0 & d & 0\\ c & 0 & 0 \end{vmatrix}.$$ Hence, $$A=0,\quad B=-c^2,\quad C=d^2,\quad F=0,\quad G=-cd,\quad H=0.$$ Therefore, the reciprocal cone in the translated system is $$-c^2Y^2+d^2Z^2-2cdZX=0,$$ or equivalently, $$c^2Y^2-d^2Z^2+2cdZX=0.$$ Returning to the original coordinates, $$c^2y^2-d^2(z-d)^2+2cd(z-d)x=0,$$ which may also be written as $$c^2y^2=d(z-d)(dz-2cx-d^2).$$ Hence, the required reciprocal cone has been obtained. $\blacksquare$